In Jackson's Classical Electrodynamics, he gives the Green's function for a conducting sphere with Dirichlet boundary conditions as $$ G(\mathbf{x},\mathbf{x}^\prime) = \frac{1}{|\mathbf{x} - \mathbf{x}^\prime|} - \frac{a}{x^\prime\left|\mathbf{x} - \frac{a^2}{x^{\prime2}}\mathbf{x}^\prime\right|}. $$ The way this Green function was obtained was from using the method of images on a grounded (zero potential at the surface) conducting sphere with a point charge outside the sphere at position $\mathbf{x}^\prime$. The second term then corresponds to the image charge.

So this Green function was obtained using a specific case, i.e. a conducting sphere with a point charge outside, and with zero potential at the surface. What I understand is that the Green function is specific to the boundary conditions imposed. So I would imagine that if a different boundary condition at the surface of the sphere is imposed (e.g. a constant non-zero potential at the surface of the sphere), a different Green function would correspond to it, in particular one that would correspond to the potential solved by Jackson using the method of images for the case of a conducting sphere held at a constant potential $V$: $$\Phi(\mathbf{x}) = \frac{1}{4\pi\epsilon_0}\left[\frac{q}{|\mathbf{x}-\mathbf{y}|}-\frac{aq}{y\left|\mathbf{x}-\frac{a^2}{y^2}\mathbf{y}\right|}\right] +\frac{Va}{|\mathbf{x}|},$$ where $a$ is the radius of the sphere and $\mathbf{y}$ is the position of the point charge.

However, Jackson argues that the Green function above (1st equation) would correspond to a conducting sphere for any Dirichlet boundary condition. Why is this the case?


2 Answers 2


What I understand is that the Green function is specific to the boundary conditions imposed.

No. The Green function is independent of the specific boundary conditions of the problem you are trying to solve. In fact, the Green function only depends on the volume where you want the solution to Poisson's equation. The process is:

  1. You want to solve $\nabla^2V=\displaystyle-\frac{\rho}{\epsilon_0}$ in a certain volume $\varOmega$.

  2. You define the Green function as one which is a solution of ${\nabla^{\prime}}^2G(\mathbf{r},\mathbf{r^\prime})=-4\pi\delta(\mathbf{r}-\mathbf{r'})$ with null Dirichlet boundary conditions, i.e. $G(\mathbf{r},\mathbf{r'})=0$ for all $\mathbf{r'\in\partial\varOmega}$.

    NB: Be careful, in Jackson the Green function has a $4\pi$ floating around in the definition, but in other texts the $4\pi$ is absorbed in the Green function and the definition is just ${\nabla^{\prime}}^2G(\mathbf{r},\mathbf{r^\prime})=-\delta(\mathbf{r}-\mathbf{r'})$.

  3. Then construct the solution to your problem as $$V(\mathbf{r})=\frac{1}{4\pi\varepsilon_0}\int_\varOmega G(\mathbf{r},\mathbf{r}^\prime)\rho(\mathbf{r^\prime})dV'-\frac{1}{4\pi}\int_{\partial\varOmega} V(\mathbf{r'})\frac{\partial G}{\partial n'}dS'$$ where in the second integral you are adding the specific boundary conditions of your problem.

  • $\begingroup$ Thanks for the reply @Urb. The integral you've shown in 3. makes it clear that the Green function really does not depend on the specific boundary condition of the problem. So if I understand correctly, one method of obtaining a Green function relatively easily is to solve a method of images problem with point charges for a specific geometry. For example, if I want a Green function that is appropriate for semi infinite volume (e.g. $z>0$) geometry, then I just need to solve a method of images problem for a point charge with a conducting plane at $z=0$. Is this correct? $\endgroup$ Commented May 26, 2020 at 11:33
  • 1
    $\begingroup$ You're right. And in your example the conducting plane would be grounded. $\endgroup$
    – Urb
    Commented May 26, 2020 at 18:44
  • $\begingroup$ Nice answer. Gives a big picture view of what's happening with all the Green's function business. $\endgroup$
    – Tachyon209
    Commented Oct 2, 2020 at 18:18

Green's functions can be calculated once, then used repeatedly for the same configuration, but with different input functions (e.g., $\rho(\mathbf{x})$). Considering that the fundamental Green's function (cf, the table in [1]), $G_F$, is $$ G_F(\mathbf{x};\mathbf{x}^\prime) = \frac{-1}{4\,\pi\| \mathbf{x} \|} \,, $$ I consider \begin{equation} \tag{1} G(\mathbf{x};\mathbf{x}^\prime) = \frac{-1}{4\,\pi\| \mathbf{x} - \mathbf{x}^\prime\|} + \frac{k_{1 }}{\|k_{2 }\mathbf{x} - k_{3 }\mathbf{x}^\prime\|} \,, \end{equation} where the consants obey the restriction that $ k_{2 } \neq k_{3 }\,.$ Now, I demonstrate that there exists some $k_{1 } $, $k_{2 }$, and $k_{3 }$ such that the function in (1) satisfies the four requirements for the calculation of a Green' function [2].

First Requirement

The first requirement is that $L[G(\mathbf{x};\mathbf{x}^\prime)]=0$ except at $$\mathbf{x}= \mathbf{x}^\prime\quad\text{and}\quad \mathbf{x}=\frac{k_{3}}{k_{2}}\,\mathbf{x}^\prime.$$

It is tedious to take the Laplacian of the fundamental Green's functions. It is no more tedius to take the Laplacian of each term of the Green's function in (1). One can take the Laplacian by hand or with a symbolic software package. Needless to say, both the fundamental Green's function and Green's function here satisfy the first requirement.

Second Requirement

The second requirement is that the Green's function satisfy linear homogeneous boundary conditions $$B_j[G(\mathbf{x};\mathbf{x}^\prime)] =0,\quad\text{for}~j=1,2,\ldots ,m\,.$$ For the purpose of calculating a Green's function, the requirements must pertain to linear homogeneous boundary conditions. These can be either homogeneous Dirichlet boundary conditions, homogeneous Neumann boundary conditions, or homogeneous mixed boundary conditions. Distinct homogeneous boundary conditions (i.e., Dirchilet, Neumann, or mixed) result in the calculation of distinct Green's functions.

In the OP's problem here, there are two boundary conditions (i.e., $B_1[G(\mathbf{x};\mathbf{x}^\prime)]$ and $B_2[G(\mathbf{x};\mathbf{x}^\prime)]$). The first boundary condition implicit in the boundary-value problem here (and implicit in many others problems) is a homogeneous Dirchilet boundary condition in the limit as $\left\|\mathbf{x} \right\| $ goes to infinity. Thus, $$B_1[G(\mathbf{x};\mathbf{x}^\prime)] = \lim_{\left\|\mathbf{x}\right\|\to\infty}G(\mathbf{x};\mathbf{x}^\prime) = 0.$$ By inspection, the fundamental Green's function as well as the Green's function in (1) meet this requirement.

The second boundary condition more or less explicit in the boundary-value problem here is a homogeneous Dirchilet boundary condition for $\left\|\mathbf{x}\right\| =a$. Thus, $$B_2[G(\mathbf{x};\mathbf{x}^\prime)] = \left.G(\mathbf{x};\mathbf{x}^\prime)\right\|_{\left\|\mathbf{x}\right\| =a} = 0.$$ By inspection, the fundamental Green's function does not meet this requirement. However, the Green's function in (1) would meet this requirement if \begin{align} \tag{2} k_1 &= \frac{1}{4\,\pi},\qquad k_2 = \frac{\left\|\mathbf{x}^\prime\right\|}{a} , \qquad k_3 = \frac{a}{\left\|\mathbf{x}^\prime\right\|}\,. \end{align} In such case (1) becomes, \begin{equation} G(\mathbf{x};\mathbf{x}^\prime) = \frac{-1}{4\,\pi\| \mathbf{x} - \mathbf{x}^\prime\|} + \frac{1}{4\,\pi\|\frac{\left\|\mathbf{x}^\prime\right\|}{a}\mathbf{x} - \frac{a}{\left\|\mathbf{x}^\prime\right\|}\mathbf{x}^\prime\|} \,, \end{equation}

Third Requirement

The third requirement is that since the Laplacian is a second-order differential operator, the $G$ must be continuous at $\mathbf{x} = \mathbf{x^\prime}$. Note that: \begin{align*} & \lim_{ \boldsymbol{\epsilon} \to\mathbf{0}} \left[\frac{-1}{4\,\pi\left\| \mathbf{x} - \mathbf{x}^\prime \right\| }\right]_{\mathbf{x} = \mathbf{x^\prime}- \boldsymbol{\epsilon}}^{\mathbf{x} = \mathbf{x^\prime} + \boldsymbol{\epsilon}} + \left[\frac{k_{1i}}{\left\| k_{2i}\,\mathbf{x} - k_{3i}\,\mathbf{x}^\prime \right\| }\right]_{\mathbf{x} = \mathbf{x^\prime}- \boldsymbol{\epsilon}}^{\mathbf{x} = \mathbf{x^\prime} + \boldsymbol{\epsilon}} \\ &\quad= \lim_{ \boldsymbol{\epsilon} \to\mathbf{0}} \left[ \frac{-1}{4\,\pi\left\| \left[ \mathbf{x^\prime}+ \boldsymbol{\epsilon}\right] - \mathbf{x}^\prime \right\| } + \frac{k_{1}}{4\,\pi\left\| \left[ \mathbf{x^\prime}- \boldsymbol{\epsilon}\right] - \,\mathbf{x}^\prime \right\| } \right] \\ &\quad+ \lim_{ \boldsymbol{\epsilon} \to\mathbf{0}} \left[ \frac{k_{1i}}{\left\| k_{2i}\,\left[ \mathbf{x^\prime}+ \boldsymbol{\epsilon}\right] - k_{3i}\,\mathbf{x}^\prime \right\| } - \frac{k_{1i}}{\left\| k_{2i}\,\left[ \mathbf{x^\prime}- \boldsymbol{\epsilon}\right] - k_{3i}\,\mathbf{x}^\prime \right\| } \right] \\ &\quad= \lim_{ \boldsymbol{\epsilon} \to\mathbf{0}} \left[ \frac{-1}{4\,\pi\left\| \boldsymbol{\epsilon} \right\| } + \frac{1}{ 4\,\pi\left\| - \boldsymbol{\epsilon} \right\| } \right] \\ &\quad+ \lim_{ \boldsymbol{\epsilon} \to\mathbf{0}} \left[ \frac{k_{1i}}{\left\| k_{2i}\, \boldsymbol{\epsilon} + \mathbf{x^\prime} \left[k_{2i} - k_{3i}\right] \right\| } - \frac{k_{1i}}{ \left\| -k_{2i}\, \boldsymbol{\epsilon} + \mathbf{x^\prime}\left[ k_{2i} -k_{3i} \right] \right\| } \right] \\ &\quad= \frac{-1}{\left\| \mathbf{x^\prime} \left[k_{2i} - k_{3i}\right] \right\| } - \frac{k_{1i}}{ \left\| \mathbf{x^\prime}\left[ k_{2i} -k_{3i} \right] \right\| } \\ &\quad= \frac{k_{1i}}{\left| k_{2i} - k_{3i}\right|\left\| \mathbf{x^\prime} \right\| } - \frac{k_{1i}}{ \left| k_{2i} - k_{3i}\right| \left\| \mathbf{x^\prime} \right\| } \\ &\quad =0\,. \end{align*} Since the evaluation of this limit is identically zero, therefore $G$ is continuous at $\mathbf{x} = \mathbf{x^\prime}$.

Fourth Requirement

The fourth requirement is that $\lim_{\boldsymbol{\epsilon}\to \mathbf{0}}\int_{\mathbf{x}^\prime - \boldsymbol{\epsilon}}^{\mathbf{x}^\prime + \boldsymbol{\epsilon}} L\left[G(\mathbf{x}-\mathbf{x}^\prime)\right]\,dV = 1$.

To evaluate this, I first must denote two geometrical objects. These are a ball and the surface of a ball. By $\mathcal{B}{\left(\mathbf{0},\epsilon\right)}$ I denote a ball with center at $\mathbf{0}$ and radius $\epsilon$. By $\mathcal{S}{\left(\mathbf{0},\epsilon\right)} $ I denote the closed spherical surface with center at $\mathbf{0}$ and radius $\epsilon$ (i..e, the boundary of the ball $\mathcal{B}{\left(\mathbf{0},\epsilon\right)}$). With these geometrical objects the fourth requirement is written as
$$\lim_{\varepsilon\to0}\iiint_{\mathcal{B}{\left(\mathbf{x}^\prime,\epsilon\right)}} L[G{\left(\mathbf{x};\mathbf{x}^\prime\right)}] \,\,d^3\mathbf{x} = 1\,.$$ Note that (1) is composed of two terms: \begin{equation} \frac{-1}{4\,\pi\| \mathbf{x} - \mathbf{x}^\prime\|} \qquad \text{and}\qquad \frac{k_{1 }}{\|k_{2 }\mathbf{x} - k_{3 }\mathbf{x}^\prime\|} \,. \end{equation} Since the evaluation of the volume integral is done only in the neighborhood around $\mathbf{x}^\prime$ and since the Laplacian of the second terms is zero in that neighborhood (see the first requirement) we have that \begin{align*} \lim_{\varepsilon\to0}\iiint_{\mathcal{B}{\left(\mathbf{x}^\prime,\epsilon\right)}} L[G{\left(\mathbf{x};\mathbf{x}^\prime\right)}] \,\,d^3\mathbf{x} = \lim_{\varepsilon\to0}\iiint_{\mathcal{B}{\left(\mathbf{x}^\prime,\epsilon\right)}} \boldsymbol{\nabla}\cdot\left[\boldsymbol{\nabla} \frac{-1}{4\,\pi\| \mathbf{x} - \mathbf{x}^\prime\|}\right] \,d^3\mathbf{x} \end{align*} Now, I do a change of variables. I write that $\boldsymbol{\chi} = \mathbf{x} - \mathbf{x}^\prime$. Thus, \begin{align*} \lim_{\varepsilon\to 0}\iiint_{\mathcal{B}{\left(\mathbf{x}^\prime,\epsilon\right)}} \boldsymbol{\nabla}\cdot\left[\boldsymbol{\nabla} \frac{-1}{4\,\pi\| \mathbf{x} - \mathbf{x}^\prime\|}\right] \,d^3\mathbf{x} &= \lim_{\varepsilon\to0}\iiint_{\mathcal{B}{\left(\mathbf{0} ,\epsilon\right)}} \boldsymbol{\nabla}\cdot\left[\boldsymbol{\nabla} \frac{-1}{4\,\pi\| \boldsymbol{\chi} \|}\right] \,d^3\boldsymbol{\chi} \end{align*} Now, I use the divergence theorem to write that \begin{align*} \lim_{\varepsilon\to 0}\iiint_{\mathcal{B}{\left(\mathbf{x}^\prime,\epsilon\right)}} \boldsymbol{\nabla}\cdot\left[\boldsymbol{\nabla} \frac{-1}{4\,\pi\| \mathbf{x} - \mathbf{x}^\prime\|}\right] \,d^3\mathbf{x} &= \lim_{\varepsilon\to0}\iint_{\mathcal{S}{\left(\mathbf{0} ,\epsilon\right)}} \left[\boldsymbol{\nabla} \frac{-1}{4\,\pi\| \boldsymbol{\chi}{\left(\rho,\theta,\varphi\right)} \|}\right] \cdot d\mathbf{S} \end{align*} I know that $\| \boldsymbol{\chi}{\left(\rho,\theta,\varphi\right)} \| = \rho$. Thus, $\boldsymbol{\nabla} \frac{-1}{4\,\pi\| \boldsymbol{\chi}{\left(\rho,\theta,\varphi\right)} \|} = \boldsymbol{\nabla} \frac{1}{4\,\pi \,\rho^2}$. Further, since I am integrating over a sphere of radius $\epsilon$, $$ \left(\boldsymbol{\nabla} \frac{-1}{4\,\pi\| \boldsymbol{\chi}{\left(\rho,\theta,\varphi\right)} \|}\right)_{\mathcal{S}{\left(\mathbf{0} ,\epsilon\right)}} = \frac{1}{4\,\pi \,\epsilon^2},$$ and the surface element, $d\mathbf{S} $, evaluated on the surface of the ball is given by the equation $$ \left(d\mathbf{S}\right)_{\mathcal{S}{\left(\mathbf{0} ,\epsilon\right)}} =\epsilon^2\,\sin\theta \,d\theta \,d\varphi \,\hat{\boldsymbol{\rho}}.$$ Consequently, \begin{align*} \lim_{\varepsilon\to 0}\iiint_{\mathcal{B}{\left(\mathbf{x}^\prime,\epsilon\right)}} \boldsymbol{\nabla}\cdot\left[\boldsymbol{\nabla} \frac{-1}{4\,\pi\| \mathbf{x} - \mathbf{x}^\prime\|}\right] \,d^3\mathbf{x} &= \frac{1}{4\,\pi} \iint_{\mathcal{S}{\left(\mathbf{0} ,\epsilon\right)}} \sin\theta \,d\theta \,d\varphi \end{align*} Finally, since $ \iint_{\mathcal{S}{\left(\mathbf{0} ,\epsilon\right)}} \sin\theta \,d\theta \,d\varphi= 4\,\pi $, we have shown that
the function in (1) satisfies the fourth requirement.

Subject to the constraints in (2), the function in (1) satisfies the four requirements necessary to construct a Green's function. Namely, the Green's function for this situation is given by the equations \begin{equation} \boxed{ G(\mathbf{x};\mathbf{x}^\prime) = \frac{-1}{4\,\pi\| \mathbf{x} - \mathbf{x}^\prime\|} + \frac{a\left\|\mathbf{x}^\prime\right\|}{4\,\pi\left\| \left\|\mathbf{x}^\prime\right\|^2 \mathbf{x} - a^2\,\mathbf{x}^\prime\right\|} \,.} \end{equation} This Green's function satisfies the specific homogeneous Dirichlet boundary conditions that are necessary for this situation. Thus, this Green's function is appropriate for this situation irrespective of the form of the input (e.g., $\rho{(\mathbf{x})}$).


[1] https://en.wikipedia.org/wiki/Green%27s_function

[2] Zwillinger, Handbook of Differential Equations, Fourth Edition, p. 210.


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